#### Answer

The solution is $x=-\dfrac{1}{4}$

#### Work Step by Step

$\dfrac{4x+3}{x+1}+\dfrac{2}{x}=\dfrac{1}{x^{2}+x}$
Take out common factor $x$ from the denominator of the fraction on the right side of the equation:
$\dfrac{4x+3}{x+1}+\dfrac{2}{x}=\dfrac{1}{x(x+1)}$
Multiply the whole equation by $x(x+1)$:
$x(x+1)\Big[\dfrac{4x+3}{x+1}+\dfrac{2}{x}=\dfrac{1}{x(x+1)}\Big]$
$(x)(4x+3)+2(x+1)=1$
Evaluate the indicated operations:
$4x^{2}+3x+2x+2=1$
Take $1$ to the left side of the equation and simplify:
$4x^{2}+3x+2x+2-1=0$
$4x^{2}+5x+1=0$
Solve by factoring:
$(x+1)(4x+1)=0$
Set both factors equal to $0$ and solve each individual equation for $x$:
$x+1=0$
$x=-1$
$4x+1=0$
$4x=-1$
$x=-\dfrac{1}{4}$
Since the original equation is undefined for $x=-1$, the solution is just $x=-\dfrac{1}{4}$